🚀 go-pugleaf

In article <30DEC199200153934@pierre.mit.edu>, chuck@pierre.mit.edu (Chuck Parsons) writes:
>   Second mechanical strength of materials is _not_ greater than the
> strength of the chemical bonds forming the material. 

This is something I wanted to address earlier - I have stated that I think
it's true, but haven't been able to prove it.

> Consider a thin hoop flywheel. Any flywheel can be thought of as being made
> up of many such hoops. Let the hoop have a fixed cross-sectional area A
> and the average mass of each atom in the material be M.
> 
>   The energy stored in each atom of the hoop is E=.5 M*V**2.
> 
>   When the hoop spins it tries to pull itself apart. It requires
>   on the order of .25*M*V**2/R  _per_ atom  of force to keep the loop intact.

That's radial stress.  The hoop stress is larger.  Force in this case on
an atom is M*V**2/D !  That's right, D, not R.  Hoop stress on a thin hoop
in rotation is much larger than radial stress.
The reasoning is as follows:  Hoop stress in the ring rho * V**2.  (rho is
density).  So the stress is M*V**2/v where v is atomic volume (D**3).
The stress is a force acting over atomic area (D**2) so force is M*V**2/D.

The remainder of my analysis differs from yours.  The total flywheel energy
is pi*M*V**2*R/D.  Thus the force on a single bond in terms of total
flywheel energy is Ef/pi*R.  The bond force is Ebond/D.  The total chemical
binding force is Ec=2*pi*R*Ebond/D.  So the bond force Ec/2*pi*R.  Then
to keep flywheel force below bond force, Ef < Ec/2.  In other words, the
rotor can only store 1/2 the energy required to disintegrate the material.

>    Since for really bitchin materials the heat of vaporization
> is similar the the binding energy, and because real materials have
> defects and fall apart well before the strain (or is it stress)
> aproaches the atomic limit. The flywheel will at _most_ have
> enough energy to vaporize itself, probably much less. The flywheel
> and enclsoure (assuming the enclosure mass is greater thant the flywheel)
> only has to get hot it doesn't have to vaporize.

Ok, now let's take a reasonable bond, C-C, for example, which has a
strength of 400 kJ/mole (33 kJ/g).  Now let's look at combustion of gasoline:
48 kJ/g.  Thus, the flywheel could store at most 16 kJ/g, and a flywheel
storing the energy equivalent of 2 gallons of gas would have a minimum mass of
51 kg.

But how close is this to real materials?  It was posted that carbon-fiber
composite had a strength of 1 Mpsi.
Density of graphite is 2.267 g/cc.  This leads to D=2.06E-10 m.
The energy per atom divided by D is 3.2E-9 N.  The stress is then
the number of atoms in a square meter times this:  7.54E10 N/m^2.
Converting to psi, that's 11 Mpsi.  So real materials only get about
1/10 th or less strength when compared to the bond energy present.
So a 510 kg flywheel is the minimum feasible flywheel to store that
2 gallons of gas energy.  Now figure shielding of at least that much
mass, and you have over a ton of material.  And you have yet to install
the electric motor, and left no room for error.

How big is that wheel?  Let's say you design it as a cylinder (preserves
the stress and energy assumptions above - if not a cylinder, it stores
less energy per mass and experiences higher stress).  Let's say the walls
are 1 cm thick.  Assume the density is the same as graphite.  Assume it
is 2 flywheels, each the length of the car, and the car is 4 m long.
Then flywheel radius for the 510 kg rotor is 44 cm (the rotors are nearly
a meter wide).  These would fit in the area of the car (two by four meters,
roughly).  But then the passengers, etc., would be 1 meter up, plus shielding
plus road clearance.  Can you see climbing up 5 or 6 feet to get into your car?

But it is realistic - I will grant it is possible, would not explode,
but will still be dangerous.  It may be possible to fully shield, but now
I'm worried about a ton of careening composite tossing my body to the
four winds or spreading it thinly across the asphalt.  These suckers will
not vaporize on failure, but rather are likely to conserve a great deal
of the energy as angular momentum, transferred to the still solid housing,
that will then transfer itself whereever it pleases.


--
Richard Stead
Center for Seismic Studies
Arlington, VA
stead@seismo.css.gov